Transformation of an $L^1$ function by a homomorphism of measure space is also an $L^1$ function.

Question

Let $(X,\mathscr{B},\mu)$ and $(Y,\mathscr{D},\nu)$ be two complete measure spaces and $\alpha:\mathscr{D}\rightarrow \mathscr{B}$ be a homomorphism, i.e., a map satisfying

1. $\alpha(A_1\cup A_2)=\alpha(A_1)\cup \alpha(A_2)$ for all $A_1,A_2\in\mathscr{D}$,
2. $\alpha(A^c)=\alpha(A)^c$ for all $A\in\mathscr{D}$,
3. $\mu(\alpha(A))=\nu(A)$ for all $A\in\mathscr{D}$.

Question: Show that if $f\in L^1(Y,\mathscr{D},\nu)$ and $g : X \to \mathbb R$ satisfy $\alpha(f^{-1}B)=g^{-1}(B)$ for all measurable $B \subseteq \mathbb R$, then $g\in L^1(X,\mathscr{B},\mu)$ and $\left\lVert f\right\rVert_1=\left\lVert g\right\rVert_1$.

If $\alpha=T^{-1}$ for some measure-preserving map $T:X\rightarrow Y$, then the answer of the above question follows from the change of variable formula for integration. But how to proceed for a general homomorphism? Thanks in advance for any help.

Answer 1

For $\lambda \ge 0$, let $B_\lambda = (-\infty,-\lambda] \cup [\lambda,\infty) \subset \mathbb{R}$. Then, $$||g||_1 = \int_0^\infty \mu(g^{-1} B_\lambda)d_\lambda = \int_0^\infty \mu(\alpha(f^{-1}B_\lambda))d\lambda = \int_0^\infty \nu(f^{-1}B_\lambda)d\lambda = ||f||_1.$$